NCERT Solutions for Class 9 Science Exploration Chapter 5

A mixture is a combination of two or more substances in which each component keeps its own properties.
Mixtures can be separated by using differences in solubility, boiling point, density, particle size, magnetism, or movement through a medium.

NCERT Solutions for Class 9 Science Exploration Chapter 5 help students understand Exploring Mixtures and their Separation, a Chemistry chapter from the 2026-27 Class 9 Science textbook. This chapter explains why salt dissolves in water, why muddy water settles, why milk shows light scattering, and how mixtures are separated in laboratories, industries, hospitals, farms, and homes. CBSE students should focus on the difference between solutions, suspensions and colloids, along with separation methods such as crystallization, distillation, paper chromatography, separating funnel, sublimation, centrifugation and coagulation. These NCERT Class 9 Science Solutions cover in-text Pause and Ponder questions and all Class 9 Science Chapter 5 exercise solutions.

Key Takeaways

• Mixture types: Mixtures may be homogeneous, heterogeneous, solutions, suspensions, or colloids.
• Concentration: A solution’s concentration can be expressed as % m/m, % m/v, or % v/v.
• Separation basis: Separation methods depend on solubility, boiling point, density, particle size, and sublimation.
• Light scattering: Tyndall effect is shown by colloids and suspensions, but not by true solutions.

NCERT Solutions for Class 9 Science Exploration Chapter 5 Structure 2026

NCERT Solutions for Class 9 Science Exploration Chapter 5 In-Text Questions

Class 9 Science Exploration Chapter 5 uses everyday examples like ORS, vinegar, blood, salt crystals, muddy water, milk, and perfume-making to explain mixtures. These answers follow the textbook’s concepts and use direct, exam-ready explanations.

Q1. How much zinc oxide is present in 300 g of talcum powder containing 4% m/m zinc oxide?

Answer: The talcum powder contains $12 , g$ of zinc oxide.

Explanation:
Given:

$Mass , of , talcum , powder = 300 , g$

$Zinc , oxide = 4% , m/m$

Formula:

$Mass , of , solute = \frac{Percentage \times Mass , of , mixture}{100}$

$Mass , of , zinc , oxide = \frac{4 \times 300}{100}$

$Mass , of , zinc , oxide = 12 , g$

So, $300 , g$ of talcum powder contains $12 , g$ zinc oxide.

Q2. Find the % v/v of orange juice concentrate in the prepared drink.

Question: Two tablespoons of orange juice concentrate are mixed with water to make 150 mL juice. Each tablespoon is 15 mL. Find the % v/v of concentrate.

Answer: The orange juice concentrate is $20% , v/v$.

Explanation:
Volume of concentrate:

$2 \times 15 = 30 , mL$

Total volume of juice:

$150 , mL$

Formula:

$Volume , by , volume , percentage = \frac{Volume , of , solute}{Volume , of , solution} \times 100$

$= \frac{30}{150} \times 100$

$= 20% , v/v$

The juice contains $20% , v/v$ orange concentrate.

Q3. How will you prepare vinegar from glacial acetic acid?

Answer: Vinegar can be prepared by diluting glacial acetic acid to make a $5% , v/v$ acetic acid solution.

Explanation:
Glacial acetic acid is $100%$ acetic acid. Vinegar contains $5% , v/v$ acetic acid.

To prepare $100 , mL$ vinegar:

1. Take $5 , mL$ glacial acetic acid.
2. Add water carefully to make the final volume $100 , mL$.
3. Mix the solution well.

This gives $5% , v/v$ acetic acid solution.

Safety note: Glacial acetic acid is corrosive and should be handled only under teacher supervision.

Q4. From solubility curves, which compound deposits more solid on cooling from 80 °C to 60 °C?

Answer: Compound B is likely to deposit more solid if its solubility decreases more sharply between 80 °C and 60 °C.

Explanation:
A hot saturated solution contains the maximum solute dissolved at that temperature. When it cools, solubility decreases.

The extra solute that can no longer remain dissolved separates as crystals.

To decide which solution deposits more solid, compare the fall in solubility from 80 °C to 60 °C.

The compound with the larger decrease in solubility deposits more crystals.

Q5. Will crystal size change if evaporation rate is increased or decreased?

Answer: Yes, slower evaporation usually forms larger and better-shaped crystals, while faster evaporation forms smaller crystals.

Explanation:
Slow evaporation gives particles more time to arrange in a regular pattern. This helps form larger and well-shaped crystals.

Fast evaporation removes solvent quickly. Particles get less time to arrange properly.

So, rapid evaporation often produces smaller and less regular crystals.

Q6. State whether the chromatography and separation statements are true or false.

Answer: The corrected statements are given below.

Explanation:
Evaporation is used when only the solute is required. Distillation is used when the liquid component also needs to be recovered.

Paper chromatography separates components based on their movement with the solvent on paper.

Q7. Why do immiscible liquids form two separate layers in a separating funnel?

Answer: Immiscible liquids form separate layers because they do not mix and have different densities.

Explanation:
Oil and water do not mix because their particles do not interact uniformly. The liquid with lower density forms the upper layer.

In a mustard oil and water mixture, mustard oil forms the upper layer because it is less dense than water.

A separating funnel Class 9 question uses this difference in density to separate two immiscible liquids.

Q8. Is sublimation different from evaporation?

Answer: Yes, sublimation and evaporation are different processes.

Explanation:
Sublimation is the direct change of a solid into vapour without becoming liquid. Camphor and naphthalene show sublimation.

Evaporation is the change of a liquid into vapour from its surface. Water evaporating from a wet cloth is an example.

So, sublimation Class 9 Science questions involve solid-to-vapour change, while evaporation involves liquid-to-vapour change.

Q9. What type of mixture are clouds?

Answer: Clouds are colloids because tiny water droplets or ice crystals are dispersed in air.

Explanation:
Cloud particles are too small to settle quickly but large enough to scatter light. This matches the behaviour of a colloid.

In clouds:

1. Dispersed phase = tiny water droplets or ice crystals
2. Dispersion medium = air

Clouds show light scattering, so they are not true solutions.

Q10. Why do cities with smoke and dust often look hazy?

Answer: Cities with smoke and dust look hazy because suspended particles scatter light.

Explanation:
Smoke and dust particles in air scatter sunlight in different directions. This scattering makes the path of light visible and reduces clarity.

This is an example of the Tyndall effect Class 9 concept. Suspensions and colloids show the Tyndall effect because their particles are large enough to scatter light.

NCERT Solutions for Class 9 Science Exploration Chapter 5 Exercise Questions

The Revise, Reflect, Refine section includes MCQs, numericals, table completion, separation-method reasoning, and solubility-based questions. These Class 9 Science Chapter 5 exercise solutions are written in direct answer format.

Q1. Which mixtures are correctly classified as homogeneous and heterogeneous?

Answer: The correct option is (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm.

Explanation:
Muddy water is heterogeneous because solid particles are suspended in water.

Milk and blood are colloids, so they are treated as heterogeneous mixtures even though they may look uniform.

Brass is an alloy and appears uniform throughout, so it is a homogeneous mixture.

Option (iv) gives the correct classification.

Q2. Which mixtures show the Tyndall effect?

Answer: The correct option is (iii) a and c.

Explanation:
The mixtures are:

(a) Air and dust particles
(b) Copper sulfate and water
(c) Starch and water
(d) Acetone and water

Air with dust particles scatters light. Starch in water forms a colloidal mixture and also scatters light.

Copper sulfate solution and acetone-water mixture are true solutions. They do not show the Tyndall effect.

So, the correct answer is a and c.

Q3. Complete the table for solution, suspension, and colloid.

Answer: The completed table is given below.

Explanation:
Solutions suspensions and colloids Class 9 questions depend mainly on particle size, settling, filtration, and Tyndall effect.

Solutions have the smallest particles. Suspensions have the largest particles. Colloids have intermediate-sized particles.

Q4. Solve the concentration problems.

Q4(i). Express concentration of dry ingredients in the cake mixture.

Answer: Sugar is $15% , m/m$, flour is $84% , m/m$, and sodium hydrogencarbonate is $1% , m/m$.

Explanation:
Masses:

$Sugar = 75 , g$

$Flour = 420 , g$

$Sodium , hydrogencarbonate = 5 , g$

Total mass:

$75 + 420 + 5 = 500 , g$

Formula:

$Mass , percentage = \frac{Mass , of , component}{Total , mass} \times 100$

Sugar:

$\frac{75}{500} \times 100 = 15%$

Flour:

$\frac{420}{500} \times 100 = 84%$

Sodium hydrogencarbonate:

$\frac{5}{500} \times 100 = 1%$

Q4(ii). Calculate copper and zinc in 120 g brass containing 70% copper.

Answer: The brass contains $84 , g$ copper and $36 , g$ zinc.

Explanation:
Mass of brass:

$120 , g$

Copper:

$\frac{70}{100} \times 120 = 84 , g$

Zinc percentage:

$100 - 70 = 30%$

Zinc:

$\frac{30}{100} \times 120 = 36 , g$

So, $120 , g$ brass contains $84 , g$ copper and $36 , g$ zinc.

Q5. Cooking oil pack says one litre is 910 g. Will oil form a separate layer with water?

Answer: Yes, cooking oil will form a separate layer above water.

Explanation:
One litre of water has a mass of about $1000 , g$, while one litre of this oil has a mass of $910 , g$.

Oil is less dense than water. It does not mix with water, so it forms the upper layer.

The two layers can be separated using a separating funnel.

Steps:

1. Pour the oil-water mixture into a separating funnel.
2. Allow it to stand undisturbed.
3. Open the stopcock to drain the lower water layer.
4. Close the stopcock before the oil layer enters the flask.
5. Collect the oil separately.

The apparatus includes a separating funnel, stopcock, stand, and conical flask.

Q6. Assertion and Reason: Solutions do not exhibit the Tyndall effect.

Answer: The correct option is (iii) A is true, but R is false.

Explanation:
Assertion is true because true solutions do not show Tyndall effect.

Reason is false because particles in solutions are not larger than 100 nm. They are smaller than 1 nm.

Solution particles are too small to scatter light. So, the light path is not visible in a true solution.

Q7. How would you separate the given mixtures?

Answer: The separation methods are given below.

Explanation:
Separation of mixtures Class 9 questions depend on the physical property that differs between components.

Density, solubility, boiling point, particle size, and sublimation are common separation bases.

Q8. How can miscible liquids A and B be separated if their boiling points are 60 °C and 90 °C?

Answer: Liquids A and B can be separated by distillation.

Explanation:
The boiling point difference is:

$90°C - 60°C = 30°C$

Distillation Class 9 Science applies when two miscible liquids have a boiling point difference of about $25°C$ or more.

Liquid A boils first at $60°C$. Its vapours are cooled in the condenser and collected as distillate.

Liquid B remains in the distillation flask because it has a higher boiling point.

The labelled diagram should include:

1. Distillation flask
2. Thermometer
3. Water condenser
4. Water inlet
5. Water outlet
6. Receiver or conical flask
7. Burner

Q9. Compare evaporation, crystallization, and distillation.

Answer: Evaporation, crystallization, and distillation are used in different separation situations.

Explanation:
Evaporation can give salt from salt solution, but water is lost.

Crystallization gives purer and better-shaped crystals.

Distillation recovers the liquid component and can separate miscible liquids with different boiling points.

Q10. Blood is a colloidal mixture. What if blood behaved like a true suspension?

Answer: If blood behaved like a true suspension, its cells would settle inside blood vessels and disturb circulation.

Explanation:
Suspension particles settle when left undisturbed. If blood cells settled like suspension particles, oxygen transport and nutrient transport would be affected.

Blood behaves as a colloid because its components remain dispersed enough to circulate.

In blood:

1. Dispersed phase = blood cells and platelets
2. Dispersion medium = plasma

This makes blood a colloidal mixture.

Q11. Separate a mixture of sand, common salt, and naphthalene.

Answer: The correct sequence is sublimation, dissolution, filtration, and evaporation.

Explanation:
Step 1: Heat the mixture gently.
Naphthalene sublimes and separates from sand and salt.

Step 2: Add water to the remaining sand and salt mixture.
Common salt dissolves in water, but sand does not.

Step 3: Filter the mixture.
Sand remains as residue on filter paper.

Step 4: Evaporate the filtrate.
Common salt is obtained from salt solution.

Correct sequence:

$Sublimation \rightarrow Dissolution \rightarrow Filtration \rightarrow Evaporation$

Q12. Why is distillation effective for separating water and acetone?

Answer: Distillation is effective because acetone and water are miscible liquids with sufficiently different boiling points.

Explanation:
Acetone boils at about $56°C$ and water boils at $100°C$.

The boiling point difference is large enough for acetone to vaporise first. Its vapours can be condensed and collected separately.

Water remains in the distillation flask for longer because it has a higher boiling point.

Q13. Answer using the solubility data.

Q13(i). What mass of potassium nitrate is needed for saturated solution in 50 g water at 40 °C?

Answer: $31 , g$ potassium nitrate is needed.

Explanation:
At $40°C$, solubility of potassium nitrate is:

$62 , g$ per $100 , g$ water

For $50 , g$ water:

$Mass = \frac{62}{100} \times 50$

$Mass = 31 , g$

So, $31 , g$ potassium nitrate is needed.

Q13(ii). What happens when saturated potassium chloride solution at 80 °C cools to room temperature?

Answer: Potassium chloride crystals will separate out as the solution cools.

Explanation:
At $80°C$, solubility of potassium chloride is $54 , g$ per $100 , g$ water.

At about $25°C$, its solubility is lower than at $80°C$.

The extra dissolved potassium chloride cannot remain in solution. It separates as crystals.

Q13(iii). What is the effect of temperature on solubility of the salts?

Answer: The solubility of most given salts increases with temperature, but the increase is different for each salt.

Explanation:
From $10°C$ to $80°C$:

1. Potassium nitrate increases from $21$ to $167 , g$ per $100 , g$ water.
2. Sodium chloride increases slightly from $36$ to $37 , g$ per $100 , g$ water.
3. Potassium chloride increases from $35$ to $54 , g$ per $100 , g$ water.
4. Ammonium chloride increases from $24$ to $66 , g$ per $100 , g$ water.

Potassium nitrate shows the greatest increase. Sodium chloride shows the least change.

Q14. Calculate sugar concentration for three students.

Q14(i). Calculate % m/m concentration.

Answer: Student A has $20%$, Student B has $16.67%$, and Student C has $27.27%$ sugar solution.

Explanation:

For Student A:

$Sugar = 20 , g$

$Water = 80 , g$

$Solution = 20 + 80 = 100 , g$

$% , m/m = \frac{20}{100} \times 100 = 20%$

For Student B:

$Sugar = 20 , g$

$Water = 100 , g$

$Solution = 120 , g$

$% , m/m = \frac{20}{120} \times 100 = 16.67%$

For Student C:

$Sugar = 30 , g$

$Water = 80 , g$

$Solution = 110 , g$

$% , m/m = \frac{30}{110} \times 100 = 27.27%$

Q14(ii). Whose solution is most concentrated?

Answer: Student C’s solution is the most concentrated.

Explanation:
Student C has the highest mass percentage of sugar.

$27.27% > 20% > 16.67%$

So, Student C’s solution contains the largest amount of sugar per $100 , g$ of solution.

Q15. Identify the separation technique and suitable mixtures.

Answer: The technique marked S is distillation.

Explanation:
The apparatus labels are:

A: Distillation flask
B: Condenser
C: Receiver or conical flask

Distillation separates miscible liquids when their boiling points differ sufficiently. It can also separate a liquid from a dissolved solid.

Using the boiling point data:

Simple distillation is suitable when boiling point difference is about $25°C$ or more.

NCERT Solutions for Class 9 Science Exploration

Topics Covered in NCERT Solutions for Class 9 Science Exploration Chapter 5

Class 9 Science Exploration Chapter 5 covers mixture classification and separation techniques. The chapter connects laboratory methods with real-life uses in medicine, food preservation, farming, perfumes, water purification, and recycling.

• Homogeneous and heterogeneous mixtures Class 9
• Solutions, suspensions and colloids Class 9
• Solute and solvent
• Concentration of solution Class 9
• Mass by mass percentage
• Mass by volume percentage
• Volume by volume percentage
• Solubility and saturated solution
• Solubility curve
• Crystallization Class 9 Science
• Salt crystals from seawater
• Distillation Class 9 Science
• Fractional distillation
• Paper chromatography Class 9
• Separating funnel Class 9
• Sublimation Class 9 Science
• Alloys as homogeneous mixtures
• Centrifugation Class 9
• Coagulation Class 9 Science
• Colloids and emulsions
• Tyndall effect Class 9
• Sewage treatment and waste segregation

Important Concepts in NCERT Solutions for Class 9 Science Exploration Chapter 5

Exploring Mixtures and their Separation Class 9 depends on understanding the property used in each separation method. Students should connect every method with the reason it works.

Important Formulas in Class 9 Science Exploration Chapter 5

Class 9 Science Chapter 5 solutions include percentage concentration numericals. These formulas must be written with correct mass and volume units.